Wykorzystujemy tożsamość:
[tex]\sin\alpha=2\sin\dfrac{\alpha}{2}\cos\dfrac{\alpha}{2}[/tex]
oraz:
[tex]\sin^2\dfrac{\alpha}{2}+\cos^2\dfrac{\alpha}{2}=1[/tex]
Korzystamy z pierwszej tożsamości:
[tex]2\sin\dfrac{\alpha}{2}\cos\dfrac{\alpha}{2}=\dfrac{3}{5}\\\\\\\sin\dfrac{\alpha}{2}\cos\dfrac{\alpha}{2}=\dfrac{3}{10}\\\\\\\cos\dfrac{\alpha}{2}=\dfrac{3}{10\sin\frac{\alpha}{2}}[/tex]
Wstawiamy to do drugiej tożsamości:
[tex]\sin^2\dfrac{\alpha}{2}+\dfrac{9}{100\sin^2\frac{\alpha}{2}}=1\\\\\\100\sin^4\dfrac{\alpha}{2}-100\sin^2\dfrac{\alpha}{2}+9=0[/tex]
Wprowadzamy oznaczenie:
[tex]t=\sin^2\dfrac{\alpha}{2}[/tex]
Wówczas:
[tex]100t^2-100t+9=0\\\\a=100,\ b=-100,\ c=9\\\\\Delta=b^2-4ac=(-100)^2-4\cdot100\cdot9=10000-3600=6400\\\\\sqrt{\Delta}=80\\\\t_1=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{100-80}{200}=0,1\\\\t_2=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{100+80}{200}=0,9[/tex]
Wobec tego mamy:
[tex]\sin^2\dfrac{\alpha}{2}=\dfrac{1}{10}\quad\text{lub}\quad \sin^2\dfrac{\alpha}{2}=\dfrac{9}{10}\\\\\\\Big(\sin\dfrac{\alpha}{2}=-\dfrac{1}{\sqrt{10}}\ \text{ lub }\ \sin\dfrac{\alpha}{2}=\dfrac{1}{\sqrt{10}}\Big)\ \text{lub}\ \Big(\sin\dfrac{\alpha}{2}=-\dfrac{3}{\sqrt{10}}\ \text{ lub }\sin\dfrac{\alpha}{2}=\dfrac{3}{\sqrt{10}}\Big)[/tex]
[tex]\boxed{\sin\dfrac{\alpha}{2}=-\dfrac{3\sqrt{10}}{10}\quad\text{lub}\quad\sin\dfrac{\alpha}{2}=-\dfrac{\sqrt{10}}{10}\quad\text{lub}\quad\sin\dfrac{\alpha}{2}=\dfrac{\sqrt{10}}{10}\quad\text{lub}\quad \sin\dfrac{\alpha}{2}=\dfrac{3\sqrt{10}}{10}}[/tex]
